Question

5 A rack has 5 different pairs of shoes. The number
of ways in which 4 shoes can be chosen from it
so that there will be no complete pair, is
a 1920
b 200
C 110
d 80​

Answer 1

The number  of ways in which 4 shoes can be chosen from it , so that there will be no complete pair is 80 ways

Solution:

Given that a rack has 5 different pairs of shoes

We have to find the number  of ways in which 4 shoes can be chosen from it

, so that there will be no complete pair. So we have to use "combinations"

The formula for combination is given as:

$^{n} C_{r}=\frac{n !}{(n-r) ! r !}$

where n! means the factorial of n

$\text { For example } 3 !=3 \times 2 \times 1=6$

First of all we have to choose 4 racks out of 5 which can be chosen by:

$\begin{array}{l}{^{5} \mathrm{C}_{4}=\frac{5 !}{(5-4) ! 4 !}} \\\\ {=\frac{5 !}{4 !}=\frac{5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}=5}\end{array}$

Then from each rack we have to choose 1 shoes ,

So out of 2 we have to choose one which can be done as follows  

$=^{2} \mathrm{C}_{1} \times^{2} \mathrm{C}_{1} \times^{2} \mathrm{C}_{1} \times^{2} \mathrm{C}_{1}=16$

Total number of ways of choosing are $16 \times 5 = 80ways$

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